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6

Wind instruments work by setting up sanding waves in the air column inside them. Shorter instruments have shorter air columns and thus standing waves with shorter wavelengths resulting in higher pitches.


6

From the Wikipedia article on sound: In physics, sound is a vibration that propagates as a typically audible mechanical wave of pressure and displacement. To fully understand how is air vibrating in an open pipe, you have to consider not only the acoustic pressure wave, $$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial ...


5

What you are hearing is mains hum: mains electricity is alternating current (ie the voltage is approximately sinusoidal and symmetric about zero), with a frequency of 50Hz or 60Hz. things like kettles and heaters use a lot of power and parts of them will mechanically change shape at this frequency, which is audible. This kind of physical noise from things ...


5

No, the frequency will not change. If the wind is blowing at constant speed and the distance between source and observer remains constant, then the time it takes for a sound wave to get from source to observer will be constant. So the time interval between wave peaks (period T) when they are detected by the observer remains equal to the interval between ...


4

You're wondering why pressure nodes form at an open end of a tube. The answer is, they don't! It's just a reasonably good approximation. Physically, consider the air molecules at the center of the tube. Since they're far away from the edges, there's no way for them to "know" exactly when the tube ends, so the sound wave must "leak out" slightly. The ...


3

For small enough amplitudes, the speed of sound is independent of how loud the sound is. It is also true that for a wide range of frequencies, the speed of sound doesn't vary with the pitch. When you move to large amplitudes (the assumptions of linear material are challenged) and high frequencies (when the wavelength of the sound is comparable to the spacing ...


3

That is because of the termination impedances at the pipe end and mouth. The above described relationship for fundamental frequency is given for zero termination impedances (ideally open pipe) which is not the real case. The simplest way to account these impedances is to introduce corresponding length corrections, so for the $f_0$: $$ f_0 = ...


3

This is a common misconception about what boundary conditions do and how they do it (for example here). You discussed two types of boundary conditions, Neumann and Dirichlet. In Neumann boundary conditions, we impose that the derivative of the variable normal to the boundary is specified, generally to be zero. With Dirichlet, we impose the value that the ...


2

It depends: 1) On whether you mean mass density or particle density, 2) on the temperature and type of gas. The main observation is that speed of sound is approximately independent of particle density, and mostly a function of temperature. 1) The speed of sound (squared) is given by the compressibility at constant entropy per particle $$ c_s^2 = \left. ...


2

No, it is not true that sound travels faster in denser media. In fact it travels slower. In the adiabatic approximation we assume that the portions of the gas vibrate so fast that it is not able to exchange heat with the surroundings. The the longitudinal displacements can be shown to satisfy the wave equation $$\frac{\partial^2 u(x,t)}{\partial ...


2

There are many reasons why the answer to this question is quite complicated - but I am pretty sure that you are expected to treat the propagation of the sound as though it's spherical (I deduce this from the "assume the horn is a point source" instruction). The fact that the train is probably running on the ground means that you really only have a ...


2

Clouds can form at a temperature minimum. Above the clouds the temperature may increase sharply. The sound speed increases $T^{1/2}$, so immediately above a temperature inversion, there can be a region of decreasing (sound) refractive index. This can have the effect of bending sound waves back towards the Earth. The phenomenon is more normally noticed on ...


2

If the source were truly point-like, yes. But nothing is truly point-like, so no: your formula for $I(r)$ is modified for short distances (where the inner structure of the source becomes relevant). This means that, if your source is a set of speakers, then $I(r)\propto R^{-2}$ is only valid for $R\gg\ell$, where $\ell$ is, say, the radius of the diaphragm of ...


2

The speed of sound is greater in water than in air, so the wavelength in water is greater than in air. In effect the refractive index of the water is less than the refractive index of the air. For light it is generally true that the refractive index is higher for denser materials, and this is because light interactions mainly with the electrons in a medium ...


2

Adiabatic bulk modulus of air $=1.4\times 10^5$ Pa and Young's modulus of steel $=1.8 \times 10^{11}$ Pa. Density of air $= 1.2 $ kg m$^{-3}$ and of steel $8050$ kg m$^{-3}$. The interaction between the atoms within steel is via the bonds whereas the interaction in air are by molecules colliding with one another and limited by the speed at which the ...


2

There is no such thing as "infinite hertz". The equation to calculate the doppler shifted frequency breaks down when the emitter travels at, or faster than, the speed of sound. What actually happens when an object travels that fast is the sound waves all pile together and form a shock wave. This is what creates the sonic boom.


2

I know that speed of sound is independent of pressure No, it isn't. It's just not directly dependent on it so it's said that the speed of sound in gases depends on the temperature, molecular composition, and heat capacity ratio of the gas. But temperature and heat capacity ratio depend on pressure. Furthermore, the dependencies on temperature don't ...


2

When working with waves the wavelength and frequency (pitch) are inversely related. Sound waves have the relation frequency times wavelength equal the speed of sound. Wind instruments are using the longitudinal dimension of air in the instrument as a medium like a "string". In both the string and wind instruments, shorter wavelengths in the excited medium ...


1

If you are to say: ignore that there exists no medium for the sound wave to travel and calculate the time it would take for a sound wave in some medium (air, for instance) to travel that distance. It would be calculated as follows: Speed of sound in air is $|v| = 343 \frac{m}{s}$ the distance between the moon and earth is about $384.4 \times 10^6$ meters. So ...


1

The expression is independent of $\sigma$ and therefore there is no higher harmonics growth etc., only decay of them. I think your exponents have $\sigma$ in them through the $x$ terms, e.g., $$ e^{-n^{2} \alpha \ x} = e^{-n^{2} \alpha \ \bar{x} \ \sigma} = e^{-n^{2} \frac{\sigma}{\Gamma}} $$ One could also find $\sigma$ from $\Gamma = \tfrac{\sigma \ ...


1

Firstly, as you show, the decibel scale is a scale used to compare two levels: the one you're measuring, and another value. The definition of the decibel scale is:$$dB = 10 \times \log_{10} \left( \frac{I}{I_{Ref}} \right)$$where $I$ is the signal/sound/power that you want to measure, and $I_{Ref}$ is the value you want to compare it to. (If you want $bel$ ...


1

The propagation of sound over distance definitely changes with weather; but perhaps not the way you think. For example, it is my experience that if you stand at the shore of a quiet lake (say 1 km across) at night, you can hear sounds from the opposite shore. This is a result of changes in the density of the air, which gives a "focusing" effect. ...


1

You can reasonably assume that whatever idea you have, it's been considered already. That's because the math is the same for all wave phenomena, and a course of classical mechanics shows you there aren't that many different ways of extracting distance information. You can have parallax measurements (measuring angle differences), and in the case of coherent ...


1

Refraction occurs because of a change of speed of propagation of the wave. When light passes from air to water it slows down, whereas when sound travels from air to water it speeds up. Therefore sound is refracted away from the normal, whereas light is refracted towards the normal.


1

Remember that noise is unwanted frequency signal superimposed into the original frequency. This means to reduce noise, you need to block the unnecessary frequency components. The effect of noise increases while increasing the amplitude because amplitude is a measure of loudness. So the amplitude (loudness) of the noise also increase as you increase the ...


1

The problem is that your point view is too "mathematical". No offence, but every acoustician would jump to the ceiling hearing "one can just erase all const, they do not change anything". Oh, they do $-$ very much! Since one of them is the sound speed... But I get it, you solve that as a mathematical problem and we are undoubtedly grateful for such people. ...


1

Energy is proportional to amplitude squared. The energy in the wave is spread out over the surface of a sphere. The area of this surface increases as the wave propagates outwards from the source and is proportional to $r^2$. So the intensity of the wave (power/area) decreases in proportion to $1/r^2$.


1

I assume you are referring to what happens when such a plane wave is propagating parallel to a solid surface. You are correct that strictly speaking, the acoustic particle velocity, like the mean flow velocity, must obey the no-slip condition at the wall. Viscous effects become important very close to the wall only, such that the acoustic motions are no ...


1

The attenuation of sound in air is a function of frequency - the higher the frequency, the greater the attenuation per unit length. There is also a natural limit to the amplitude of a sound wave: once the peak pressure is more than twice the ambient pressure, you no longer have a traditional "wave" since air pressure cannot go negative. Third - ...



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